n testing a certain kind of missile, target accuracy is measured by the average distance X (from the target) at which the missile explodes. The distance X is measured in miles and the sampling distribution of X is given by:

X 0 10 50 100
P(X) 1⁄40 1/20 1⁄10 33⁄40

Calculate the variance of this sampling distribution.

a) 27.6

b) 5138.7

c) 761.0

d) 253.7

e) 88.0

f) None of the above

Respuesta :

Answer:

[tex]Var(X)=E(X^2)-[E(X)]^2 =8505-(88)^2 =761.0[/tex]

c) 761.0

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.  

The random variable is given by this table

X | 0 | 10 | 50 | 100 |

P(X) | 1/40 | 1/20 | 1/10 | 33/40 |

In order to calculate the expected value we can use the following formula:

[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]

And if we use the values obtained we got:

[tex]E(X)=(0)*(\frac{1}{40})+(10)(\frac{1}{20})+(50)(\frac{1}{10})+(100)(\frac{33}{40})=88[/tex]

In order to find the variance, we need to find first the second moment, given by :

[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]

And using the formula we got:

[tex]E(X^2)=(0)*(\frac{1}{40})+(100)(\frac{1}{20})+(2500)(\frac{1}{10})+(10000)(\frac{33}{40})=8505[/tex]

Then we can find the variance with the following formula:

[tex]Var(X)=E(X^2)-[E(X)]^2 =8505-(88)^2 =761.0[/tex]

And then the standard deviation would be given by:

[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{761}=27.586[/tex]

So then the best answer for this case is:

c) 761.0

Answer:

(C) 761.0

Step-by-step explanation:

This is a grouped data with the following distance and frequency of occurrence

Formula for variance = (∑ fx²/mean)- (mean)²

x = distance

f = frequency  

n = sample size

x f/n        f x²         fx     fx²  

0 1/40        1  0         0     0  

10 1/20   2 100        20     200  

50 1/10    4 2500      200     10000  

100 33/40 33 10000 3300    330000  

     

         n         ∑ fx             ∑fx²  

        40         3520     340200  

Note frequency was given as a fraction of the sample size

mean = ∑fx/n

3520/40 = 88

mean = 88

Variance = (∑fx²/mean) - (mean)²  

Variance = ( 340200/40) – (88)²

Variance = 8505 – 7744

Variance = 761.0

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